Problem: $h(n) = -2n^{2}-3(g(n))$ $g(t) = -7t^{2}-6t$ $f(t) = 6t^{3}-3t^{2}-7t-3-5(h(t))$ $ h(f(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 6(0^{3})-3(0^{2})+(-7)(0)-3-5(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -2(0^{2})-3(g(0))$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = -7(0^{2})+(-6)(0)$ $g(0) = 0$ That means $h(0) = -2(0^{2})+(-3)(0)$ $h(0) = 0$ That means $f(0) = 6(0^{3})-3(0^{2})+(-7)(0)-3+(-5)(0)$ $f(0) = -3$ Now we know that $f(0) = -3$ . Let's solve for $h(f(0))$ , which is $h(-3)$ $h(-3) = -2(-3)^{2}-3(g(-3))$ To solve for the value of $h$ , we need to solve for the value of $g(-3)$ $g(-3) = -7(-3)^{2}+(-6)(-3)$ $g(-3) = -45$ That means $h(-3) = -2(-3)^{2}+(-3)(-45)$ $h(-3) = 117$